Integrand size = 29, antiderivative size = 138 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}+\frac {a b \sin ^8(c+d x)}{4 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \] Output:
1/3*a^2*sin(d*x+c)^3/d+1/2*a*b*sin(d*x+c)^4/d-1/5*(2*a^2-b^2)*sin(d*x+c)^5 /d-2/3*a*b*sin(d*x+c)^6/d+1/7*(a^2-2*b^2)*sin(d*x+c)^7/d+1/4*a*b*sin(d*x+c )^8/d+1/9*b^2*sin(d*x+c)^9/d
Time = 0.63 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-7560 a b \cos (2 (c+d x))-1260 a b \cos (4 (c+d x))+840 a b \cos (6 (c+d x))+315 a b \cos (8 (c+d x))+12600 a^2 \sin (c+d x)+3780 b^2 \sin (c+d x)-840 a^2 \sin (3 (c+d x))-840 b^2 \sin (3 (c+d x))-1512 a^2 \sin (5 (c+d x))-504 b^2 \sin (5 (c+d x))-360 a^2 \sin (7 (c+d x))+90 b^2 \sin (7 (c+d x))+70 b^2 \sin (9 (c+d x))}{161280 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]
Output:
(-7560*a*b*Cos[2*(c + d*x)] - 1260*a*b*Cos[4*(c + d*x)] + 840*a*b*Cos[6*(c + d*x)] + 315*a*b*Cos[8*(c + d*x)] + 12600*a^2*Sin[c + d*x] + 3780*b^2*Si n[c + d*x] - 840*a^2*Sin[3*(c + d*x)] - 840*b^2*Sin[3*(c + d*x)] - 1512*a^ 2*Sin[5*(c + d*x)] - 504*b^2*Sin[5*(c + d*x)] - 360*a^2*Sin[7*(c + d*x)] + 90*b^2*Sin[7*(c + d*x)] + 70*b^2*Sin[9*(c + d*x)])/(161280*d)
Time = 0.39 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 522, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^2(c+d x) \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^5 (a+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int b^2 \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^7 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (b^8 \sin ^8(c+d x)+2 a b^7 \sin ^7(c+d x)+b^6 \left (a^2-2 b^2\right ) \sin ^6(c+d x)-4 a b^7 \sin ^5(c+d x)+b^6 \left (b^2-2 a^2\right ) \sin ^4(c+d x)+2 a b^7 \sin ^3(c+d x)+a^2 b^6 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} a^2 b^7 \sin ^3(c+d x)+\frac {1}{7} b^7 \left (a^2-2 b^2\right ) \sin ^7(c+d x)-\frac {1}{5} b^7 \left (2 a^2-b^2\right ) \sin ^5(c+d x)+\frac {1}{4} a b^8 \sin ^8(c+d x)-\frac {2}{3} a b^8 \sin ^6(c+d x)+\frac {1}{2} a b^8 \sin ^4(c+d x)+\frac {1}{9} b^9 \sin ^9(c+d x)}{b^7 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]
Output:
((a^2*b^7*Sin[c + d*x]^3)/3 + (a*b^8*Sin[c + d*x]^4)/2 - (b^7*(2*a^2 - b^2 )*Sin[c + d*x]^5)/5 - (2*a*b^8*Sin[c + d*x]^6)/3 + (b^7*(a^2 - 2*b^2)*Sin[ c + d*x]^7)/7 + (a*b^8*Sin[c + d*x]^8)/4 + (b^9*Sin[c + d*x]^9)/9)/(b^7*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77
\[\frac {\frac {b^{2} \sin \left (d x +c \right )^{9}}{9}+\frac {a b \sin \left (d x +c \right )^{8}}{4}+\frac {\left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )^{7}}{7}-\frac {2 a b \sin \left (d x +c \right )^{6}}{3}+\frac {\left (-2 a^{2}+b^{2}\right ) \sin \left (d x +c \right )^{5}}{5}+\frac {a b \sin \left (d x +c \right )^{4}}{2}+\frac {a^{2} \sin \left (d x +c \right )^{3}}{3}}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x)
Output:
1/d*(1/9*b^2*sin(d*x+c)^9+1/4*a*b*sin(d*x+c)^8+1/7*(a^2-2*b^2)*sin(d*x+c)^ 7-2/3*a*b*sin(d*x+c)^6+1/5*(-2*a^2+b^2)*sin(d*x+c)^5+1/2*a*b*sin(d*x+c)^4+ 1/3*a^2*sin(d*x+c)^3)
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {315 \, a b \cos \left (d x + c\right )^{8} - 420 \, a b \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, b^{2} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/1260*(315*a*b*cos(d*x + c)^8 - 420*a*b*cos(d*x + c)^6 + 4*(35*b^2*cos(d* x + c)^8 - 5*(9*a^2 + 10*b^2)*cos(d*x + c)^6 + 3*(3*a^2 + b^2)*cos(d*x + c )^4 + 4*(3*a^2 + b^2)*cos(d*x + c)^2 + 24*a^2 + 8*b^2)*sin(d*x + c))/d
Time = 0.91 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.55 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {a b \sin ^{8}{\left (c + d x \right )}}{12 d} + \frac {a b \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a b \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {8 b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2*(a+b*sin(d*x+c))**2,x)
Output:
Piecewise((8*a**2*sin(c + d*x)**7/(105*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) + a*b*sin(c + d*x)**8/(12*d) + a*b*sin(c + d*x)**6*cos(c + d*x)**2/(3*d) + a*b*sin(c + d*x)**4*cos(c + d*x)**4/(2*d) + 8*b**2*sin(c + d*x)**9/(315*d) + 4*b**2*si n(c + d*x)**7*cos(c + d*x)**2/(35*d) + b**2*sin(c + d*x)**5*cos(c + d*x)** 4/(5*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**2*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {140 \, b^{2} \sin \left (d x + c\right )^{9} + 315 \, a b \sin \left (d x + c\right )^{8} - 840 \, a b \sin \left (d x + c\right )^{6} + 180 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{7} + 630 \, a b \sin \left (d x + c\right )^{4} - 252 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{5} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/1260*(140*b^2*sin(d*x + c)^9 + 315*a*b*sin(d*x + c)^8 - 840*a*b*sin(d*x + c)^6 + 180*(a^2 - 2*b^2)*sin(d*x + c)^7 + 630*a*b*sin(d*x + c)^4 - 252*( 2*a^2 - b^2)*sin(d*x + c)^5 + 420*a^2*sin(d*x + c)^3)/d
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {140 \, b^{2} \sin \left (d x + c\right )^{9} + 315 \, a b \sin \left (d x + c\right )^{8} + 180 \, a^{2} \sin \left (d x + c\right )^{7} - 360 \, b^{2} \sin \left (d x + c\right )^{7} - 840 \, a b \sin \left (d x + c\right )^{6} - 504 \, a^{2} \sin \left (d x + c\right )^{5} + 252 \, b^{2} \sin \left (d x + c\right )^{5} + 630 \, a b \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/1260*(140*b^2*sin(d*x + c)^9 + 315*a*b*sin(d*x + c)^8 + 180*a^2*sin(d*x + c)^7 - 360*b^2*sin(d*x + c)^7 - 840*a*b*sin(d*x + c)^6 - 504*a^2*sin(d*x + c)^5 + 252*b^2*sin(d*x + c)^5 + 630*a*b*sin(d*x + c)^4 + 420*a^2*sin(d* x + c)^3)/d
Time = 20.72 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\sin \left (c+d\,x\right )}^7\,\left (\frac {a^2}{7}-\frac {2\,b^2}{7}\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {2\,a^2}{5}-\frac {b^2}{5}\right )+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,b\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^8}{4}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^2*(a + b*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^7*(a^2/7 - (2*b^2)/7) - sin(c + d*x)^5*((2*a^2)/5 - b^2/5) + (a^2*sin(c + d*x)^3)/3 + (b^2*sin(c + d*x)^9)/9 + (a*b*sin(c + d*x)^4)/2 - (2*a*b*sin(c + d*x)^6)/3 + (a*b*sin(c + d*x)^8)/4)/d
Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right )^{3} \left (140 \sin \left (d x +c \right )^{6} b^{2}+315 \sin \left (d x +c \right )^{5} a b +180 \sin \left (d x +c \right )^{4} a^{2}-360 \sin \left (d x +c \right )^{4} b^{2}-840 \sin \left (d x +c \right )^{3} a b -504 \sin \left (d x +c \right )^{2} a^{2}+252 \sin \left (d x +c \right )^{2} b^{2}+630 \sin \left (d x +c \right ) a b +420 a^{2}\right )}{1260 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)**3*(140*sin(c + d*x)**6*b**2 + 315*sin(c + d*x)**5*a*b + 180 *sin(c + d*x)**4*a**2 - 360*sin(c + d*x)**4*b**2 - 840*sin(c + d*x)**3*a*b - 504*sin(c + d*x)**2*a**2 + 252*sin(c + d*x)**2*b**2 + 630*sin(c + d*x)* a*b + 420*a**2))/(1260*d)